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Question Number 129584 by bemath last updated on 16/Jan/21

$$\underset{x\to 0}{\lim }\frac{\left(\sin \mathrm{x}\right)\left(\arctan \mathrm{x}\right)-{\mathrm{x}}^2}{1-\cos \left({\mathrm{x}}^2\right)}=?$$

Answered by liberty last updated on 16/Jan/21

$$\underset{x\to 0}{\lim }\frac{(\mathrm{x}-\frac{{\mathrm{x}}^3}{6}+\frac{{\mathrm{x}}^5}{120})(\mathrm{x}-\frac{{\mathrm{x}}^3}{3}+\frac{{\mathrm{x}}^5}{5})-{\mathrm{x}}^2}{1-(1-\frac{{\mathrm{x}}^4}{2})}=$$$$\underset{x\to 0}{\lim }\frac{{\mathrm{x}}^2[(1-\frac{{\mathrm{x}}^2}{6}+\frac{{\mathrm{x}}^4}{120})(1-\frac{{\mathrm{x}}^2}{3}+\frac{{\mathrm{x}}^4}{5})-1]}{\frac{{\mathrm{x}}^4}{2}}$$$$\underset{x\to 0}{\lim }\frac{2(1-(\frac{{\mathrm{x}}^2}{3}-\frac{{\mathrm{x}}^4}{5})-\frac{{\mathrm{x}}^2}{6}(1-\frac{{\mathrm{x}}^2}{3}+\frac{{\mathrm{x}}^4}{5})+\frac{{\mathrm{x}}^4}{120}(1-\frac{{\mathrm{x}}^3}{3}+\frac{{\mathrm{x}}^4}{5})-1)}{{\mathrm{x}}^2}=$$$$=2(-\frac{1}{3}-\frac{1}{6})=2(-\frac{1}{2})=-1$$

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